考虑一下插⼊法
n<=100n<=100n<=100f[i][j]f[i][j]f[i][j]表⽰111~iii的全排列有j个逆序对的⽅案数f[i][j]=Σf[i−1][j−k](0<=k<=i−1)f[i][j]=Σf[i-1][j-k] (0<=k<=i-1)f[i][j]=Σf[i−1][j−k](0<=k<=i−1)O(m∗n2)O(m*n^2)O(m∗n2)拓展:如果n<=1000n<=1000n<=1000呢?
n<=1000n<=1000n<=1000?
f[i][j]f[i][j]f[i][j]是f[i−1]f[i-1]f[i−1]中连续⼀段的和 前缀和优化O(n∗m)O(n*m)O(n∗m)下面上非拓展的代码:
#includeusing namespace std;int f[105][6005];int main(){ int n,k; scanf("%d%d",&n,&k); f[1][0]=1; f[2][0]=1; f[2][1]=1; f[0][0]=1; for(int i=3;i<=n;i++) { for(int j=0;j<=k;j++) { for(int kk=0;kk<=i-1&&kk<=j;kk++) { f[i][j]+=f[i-1][j-kk]%10000; } } } printf("%d",f[n][k]%10000); return 0;}